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Fudgin [204]
4 years ago
7

A 1100 kg car pushes a 1800 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the

car push against the ground with a force of 4500 N.A) What is the magnitude of the force of the car on the truck?B) What is the magnitude of the force of the truck on the car?
Physics
1 answer:
12345 [234]4 years ago
6 0

Answer:The answer is 3000 N.

Force (F) is the multiplication of mass (m) and acceleration (a).

F = m · a

It is given:

mc = 1000 kg

mt = 2000 kg

total force: F = 4500 N 

total mass: m = mc + mt

Let's calculate acceleration which is common:

a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²

Now, when we know acceleration, let's calculate force on the truck:

Ft = mt · a = 2000 · 1.5 = 3000 N

Explanation:

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If you separate vector B into its components. How many components will it have? Those components will be called?
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The vector B will have two components and those components will be called resultant vectors.

<h3>What is a component vector?</h3>

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A vector can be represented in x - direction and y - direction.

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Thus, the vector B will have two components and those components will be called resultant vectors.

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3 0
2 years ago
A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS
Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux  GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;  

transferred energy + generated energy = 0

(radiation + convection) +  generated energy = 0

[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0

[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s

(a) the electrical power generated for still summer day

T_s = T_{\infty} = 35 ^oC = 308 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k

P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W

(b)the electrical power generated for a breezy winter day

T_s = T_{\infty} = -15 ^oC = 258 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k

P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W

3 0
3 years ago
Please help me find the answer ​
Hoochie [10]

Answer:

26b) 66.7%

27) 500 N

Explanation:

26.a) In a two pulley system, the load is attached to one of the pulleys.  The other pulley is attached to a fixed surface, as well as one end of the rope.  The other end of the rope goes around moving pulley, then around the fixed pulley.

26.b) Mechanical advantage is the ratio between the forces:

MA = load force / effort force

Efficiency is the ratio between the work:

e = work done on load / work done by effort

Work is force times distance.

e = (F load × d load) / (F effort × d effort)

Rearranging:

e = (F load / F effort) × (d load / d effort)

e = MA × (d load / d effort)

In a two pulley system, the load moves half the distance of the effort.  So the efficiency is:

e = (4/3) × (1/2)

e = 2/3

e = 66.7%

27) In a three pulley system, the load moves a third of the distance of the effort.

e = (F load / F effort) × (d load / d effort)

0.40 = (600 N / F) × (1/3)

F = 500 N

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3 years ago
6. Why does sound travel faster through a steel bar than through a swimming pool filled with water?
Lady_Fox [76]
I understand that sound travels faster in water then in air. Water is a liquid, and air is gas.

Water still has the ability to roll the molecules over each other (so water can flow), it has some flexibility.

But I do not understand how a solid that is inflexible can make sound waves travel faster then in a flexible liquid.

In fact, sound waves travel over 17 times faster through steel than through air.

Sound waves travel over four times faster in water than it would in air.

3 0
3 years ago
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